% David Johnson
% EE 590A 
%
% Assignment: research article 
% Started: 25 March 2007
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\documentclass[12pt]{iopart}
%\usepackage{epsfig}			% Adds EPS support
\usepackage{graphicx}

\begin{document}

\title[EE512 article: Solution for open-ended waveguide]{Solution for Open-Ended Waveguide Analysis of an Infinite
Dielectric Half-Space}
\author[D. Johnson]{David Johnson}


\ead{djohnso@gmail.com}

\date{6 December 2007}

\begin{abstract}
The use of open-ended rectangular waveguides in non-destructive evaluation techniques
has become a viable method for inspecting or testing in a wide variety of 
applications.  As research progresses, the use of single-sided, 
reflection-based, measurement methods show significant promise particularly
in the area of evaluating composite structures.  With the progression of such
technology, so too does desire for an analytical model for each application.
This enables better characterization of the various parameters of the 
technique and provides a rigorous base for analysis of acquired data.   
The analysis in this article considers a waveguide applied to a dielectric half-space.
\end{abstract}
					
\maketitle

\section{Introduction\label{sec:intro}}
Over the past few decades, single-sided microwave non-destructive evaluation 
(NDE) techniques have matured greatly, finding a variety of applications in multiple 
industries, ranging from concrete testing \cite{disbond_concrete_khanfar},
\cite{cracked_cement_nadakuduti} to inspecting 
carbon-loaded composite structures \cite{carbon_comp_saleh}, 
\cite{carbon_fiber_kharkovsky} to searching for cracks in metals 
\cite{stainless_steel_crack_saka}.  In these processes, a vector 
network analyzer measures the reflection coefficient of a waveguide mounted
above the sample's surface.  From there, the material's dielectric constant
is extracted.  With this data, one may quantitatively characterize a given
sample.  It is the task of this article to derive an analytical solution
to model this approach applied to an infinite dielectric half-space.



Several works have tackled similar configurations.  Using numerical techniques, 
\cite{mautz_harrington_trans} modeled a waveguide with an infinite flange directed
into an infinite half-space.  This basic situation was analytically modeled in 
\cite{yoshitomi}.  A number of other works sought analytical solutions to similar 
situations, including a waveguide directly applied to a lossy finite material 
with a conductor backing \cite{stewart}.  Going a step further in complexity,
Baker-Jarvis et al. \cite{bakerjarvis} explored a coaxial 
probe with an air gap between the probe aperture and a dielectric of 
finite thickness, terminated by either an infinite dielectric half-space or perfect 
electric conductor (PEC).  An article by Bois et al \cite{bois}, relying heavily on the same methods found in 
\cite{yoshitomi}, presents the exact situation to be considered herein.  However, 
the derivation presented in \cite{bois} lacks consistency in formulation and rigor 
in detail.  As such, this paper shall review the various issues found in 
\cite{bois}, providing a solution with greater detail and a more explicit 
description of the methodology used.

\begin{figure}
\centering
\includegraphics[width=8.5cm]{open_ended_wg_diag_wavy.jpg}
\caption{Open-ended waveguide directed toward an infinite dielectric half-space.\label{fig:waveguide_diag}}
\end{figure}

\section{Methodology of derivation\label{sec:method}}
The chosen strategy centers on the use of Fourier transforms, following much the 
same formulation as \cite{yoshitomi}.  However, before exploring said formulation, 
one needs to know the basic set of boundary conditions and governing equations.  
Since the waveguide lies flush 
with the dielectric (i.e. no air gap), the boundary conditions for this particular problem
include the waveguide's walls and the probe-dielectric interface.  It is assumed the walls 
of the waveguide, as well as the flange on the end of the waveguide, are PECs, and that the flange
extends infinitely in the $x$ and $y$ directions.  Inside the waveguide,
\begin{equation}
\bar{E}_{x,y}^{wg}=\bar{E}_{x,y}^i+\bar{E}_{x,y}^r
\end{equation}

\begin{equation}
\bar{H}_{x,y}^{wg}=\bar{H}_{x,y}^i+\bar{H}_{x,y}^r.
\end{equation}
At the probe-dielectric interface ($z=0$), the tangential components of the 
fields in each region must match, resulting in

\begin{equation}
\label{E_boundary}
\bar{E}_{x,y}^t(x,y,0) = \left \{
	\begin{array}{ll}
		\bar{E}_{x,y}^i(x,y,0) + \bar{E}_{x,y}^r(x,y,0) &	{\rm for }\; |x| \le a{\rm ,} \\
		0 & {\rm elsewhere}

	\end{array} \right.
\end{equation}



\begin{equation}
\label{H_boundary}
H_{x,y}^{t}=\frac{1}{4\pi^2}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\tilde{H}_{x,y}^t(\xi,\eta)e^{-j(\xi x + \eta y)}\,d\xi\,d\eta = H_{x,y}^{wg}(x,y,0)
\end{equation}
The tangential field component of the electric field is zero on the flange as it is assumed to be an infinite PEC; conversely, the tangential component of the magnetic field is not zero due to surface currents on the flange (i.e. the electric field's normal component is not zero).  Assuming the waveguide operates in the fundamental $TE_{10}$ mode, 
the fields incident to the aperture are described by the magnetic Hertzian 
vector \cite{yoshitomi}

\begin{equation}
\label{pot_ih}
\bar{\Pi}^{ih}(x,y,z)=\frac{A^i}{(k_o^2Z_o)}\cos a_1(x+a)e^{-jk_{10}z}\hat{a}_z
\end{equation}
with wave number $k_o = \sqrt{\mu_o\epsilon_o}$ and intrinsic impedance $Z_o = \sqrt{\mu_o/\epsilon_o}$, where $\mu_o$ and $\epsilon_o$ are the permiability and permitivity of free space (respectively). The vector potential (\ref{pot_ih}), 
when applied with
\begin{equation}
\label{E_field_pot}
\bar{E} = \bar{\nabla}(\bar{\nabla}\cdot\bar{\Pi}^e) 
	+ k_1^2\bar{\Pi}^e - j\omega\mu\bar{\nabla}\times\bar{\Pi}^h
\end{equation}

\begin{equation}
\label{H_field_pot}
\bar{H} = \bar{\nabla}(\bar{\nabla}\cdot\bar{\Pi}^h) 
	+ k_1^2\bar{\Pi}^h + j\omega\epsilon\bar{\nabla}\times\bar{\Pi}^e
\end{equation}
will yield $\bar{E}^i$ and $\bar{H}^i$.  Likewise, the fields reflected
at the waveguide aperture, $\bar{E}^r$ and $\bar{H}^r$, may be calculated 
using (\ref{E_field_pot}) and (\ref{H_field_pot}) on the vector potentials
\cite{yoshitomi}

\begin{equation}
\label{pot_re}
\bar{\Pi}^{re}(x,y,z) = \sum_{m,n=1}^{\infty} \frac{A_{m,n}^e}{k_o^2}\sin a_m(x+a) \sin b_n(y+b) e^{jk_{mn}z}\hat{a}_z
\end{equation}

\begin{equation}
\label{pot_rh}
\bar{\Pi}^{rh}(x,y,z) = \sum_{\stackrel{m,n=0}{m=n\ne0}}^{\infty} \frac{A_{m,n}^h}{k_o^2Z_o}\cos a_m(x+a) \cos b_n(y+b) e^{jk_{mn}z}\hat{a}_z
\end{equation}
with $a_m = m\pi/(2a)$, $b_n = n\pi/(2b)$, and $k_{mn} = \sqrt{k_o^2-a_m^2-b_n^2}$.  
The variables $A_{mn}^e$ and $A_{mn}^e$ are unknown coefficients.  It should be noted, 
\cite{bois} mistakenly replaced $\cos b_n(y+b)$ with $\sin b_n(y+b)$ in (\ref{pot_rh}).
Similarly, the transmitted fields are represented by the vector 
potentials \cite{bois}

\begin{equation}
\label{pot_te}
\bar{\Pi}^{te}(x,y,z) =  \frac{1}{4\pi^2}\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \frac{A^e(\xi,\eta)}{k_1^2} e^{-j(\xi x + \eta y + \zeta z)}\hat{a}_z\,d\xi\,d\eta
\end{equation}

\begin{equation}
\label{pot_th}
\bar{\Pi}^{th}(x,y,z) =  \frac{1}{4\pi^2}\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \frac{A^h(\xi,\eta)}{k_1^2Z_1} e^{-j(\xi x + \eta y + \zeta z)}\hat{a}_z\,d\xi\,d\eta
\end{equation}
where $\zeta = \sqrt{k_1^2-\xi^2-\eta^2}$, $k_1$ and $Z_1$ are the respective wave 
number and impedance of the dielectric, and $A^{e,h}(\xi,\eta)$ are the unknown spectral functions.

With the basic field formulations known, efforts may now shift toward the 
approach for finding the unknown coefficients $A_{mn}^e$ and $A_{mn}^h$.  
Because of the form of (\ref{pot_te}) and (\ref{pot_th}), the application of the 
Fourier transform allows for the solving of the spectral functions $A^{e,h}(\xi,\eta)$.
Once obtained, one may then substitute the resulting $\tilde{H}^t$ equation into 
(\ref{H_boundary}).  Due to the boundary conditions on the waveguide walls,
both sides are then multiplied by $\sin a_p(x+a)\cos b_q(y+b)$ for the
$x$ component, or $\cos a_p(x+a)\sin b_q(y+b)$ for the $y$ component, 
and integrating over the dimensions of the waveguide, resulting in two
sets of linear equations, allowing for the solution of $A_{mn}^e$ and $A_{mn}^h$. 
The rationale behind these steps was never mentioned in either of the prior works.


\section{Analytic formulation\label{sec:anform}}
In order to find $A^{e,h}(\xi,\eta)$, the tangential electric field continuity 
equation, 
\begin{equation}
\label{E_con_eq}
E_{x,y}^t (x,y,0) = \frac{1}{4\pi^2}\int_{- \infty}^{\infty} \int_{- \infty}^{\infty}\tilde{E}_{x,y}^t(\xi,\eta)e^{-j(\xi x + \eta y)}\,d\xi\,d\eta = E_{x,y}^{wg}(x,y,0)
\end{equation}
shall be used in conjunction with (\ref{E_field_pot}). For the purpose of 
simplification, the Fourier transform, given by 

\begin{equation}
\label{Fourier_xform}
\tilde{f}(\xi,\eta) = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty}f(x,y)e^{j(\xi x + \eta y)}\,dx\,dy
\end{equation}
shall be taken on both sides of (\ref{E_con_eq}). The inverse 
Fourier transform is given by

\begin{equation}
\label{inv_Fourier_xform}
f(x,y) = \frac{1}{4\pi^2}\int_{- \infty}^{\infty} \int_{- \infty}^{\infty}\tilde{f}(\xi,\eta) e^{-j(\xi x + \eta y)}\,d\xi\,d\eta.
\end{equation}
The transform of (\ref{E_con_eq}) yields $\tilde{E}_{x,y}^t (\xi,\eta)= \tilde{E}_{x,y}^{wg}(\xi,\eta)$.

Because the continuity equation only contains tangential components,
(\ref{E_field_pot}) simplifies to 
\[
	\bar{E}_{x,y} = \bar{\nabla}_{tan}(\bar{\nabla}_{tan}\cdot\bar{\Pi}^e) 
	+ k_1^2\bar{\Pi}^e - j\omega\mu\bar{\nabla}_{tan}\times\bar{\Pi}^h
\]
where $\nabla_{tan} = \partial/\partial x \hat{a}_x + \partial/\partial y \hat{a}_y$.  
Using $\nabla_{tan}$ to take the curl of both sides, one can then solve for 

\begin{equation}
\label{Ah}
A^{h}(\xi,\eta) =\frac{k_1(\xi\tilde{E}_y^{wg}(\xi,\eta)-\eta\tilde{E}_x^{wg}(\xi,\eta))}{(\xi^2+\eta^2)}.
\end{equation}
Likewise, taking the tangential divergence of (\ref{E_con_eq}) yields

\begin{equation}
\label{Ae}
A^{e}(\xi,\eta) =\frac{-k_1^2(\xi \tilde{E}_x^{wg}(\xi,\eta)+\eta \tilde{E}_y^{wg}(\xi,\eta))}{\zeta(\xi^2+\eta^2)}.
\end{equation}

Knowing $A^{e}(\xi,\eta)$ and $A^{h}(\xi,\eta)$, the equations for $\tilde{H}_x^t$ 
and $\tilde{H}_y^t$ can be found using (\ref{H_field_pot}), (\ref{pot_te}), and (\ref{pot_th}), gives
% \begin{array}{Htx}
% \tilde{H}_x^{t}(\xi,\eta) & =\frac{-(\xi\eta \tilde{E}_x^{wg}(\xi,\eta)+(k_1^2-\xi^2)\tilde{E}_y^{wg}(\xi,\eta))}{\zeta k_1^2 Z_1}\\
% 				      & = \frac{-(\xi\eta (\tilde{E}_x^{i}+\tilde{E}_x^{r})(\xi,\eta)+
% 					(k_1^2-\xi^2)(\tilde{E}_y^{i}+\tilde{E}_y^{r}(\xi,\eta))}{\zeta k_1^2 Z_1}
% \end{array}

\begin{eqnarray}
\label{Htx}
\tilde{H}_x^{t}(\xi,\eta) & =\frac{-(\xi\eta \tilde{E}_x^{wg}(\xi,\eta)+(k_1^2-\xi^2)\tilde{E}_y^{wg}(\xi,\eta))}{\zeta k_1^2 Z_1}\nonumber \\
				       & = \frac{-(\xi\eta (\tilde{E}_x^{i}(\xi,\eta)+\tilde{E}_x^{r}(\xi,\eta))+
					(k_1^2-\xi^2)(\tilde{E}_y^{i}(\xi,\eta)+\tilde{E}_y^{r}(\xi,\eta))}{\zeta k_1^2 Z_1}
\end{eqnarray}

and
\begin{eqnarray}
\label{Hty}
\tilde{H}_y^{t}(\xi,\eta) & =\frac{(k_1^2-\eta^2)\tilde{E}_x^{wg}(\xi,\eta)+\xi\eta\tilde{E}_y^{wg}(\xi,\eta)}{\zeta k_1^2 Z_1}\nonumber \\
 				      & =\frac{(k_1^2-\eta^2)(\tilde{E}_x^{i}(\xi,\eta)+\tilde{E}_x^{r}(\xi,\eta))
					+\xi\eta\tilde(\tilde{E}_y^{i}(\xi,\eta)+\tilde{E}_y^{r}(\xi,\eta))}{\zeta k_1^2 Z_1}.
\end{eqnarray}

The left-hand side of the magnetic continuity equation (\ref{H_boundary})
can now be put in terms of $E_{x,y}^{wg}$.  To fill the 
right-hand side, (\ref{H_boundary}) is multiplied by $\sin a_p(x+a)\cos b_q(y+b)$ 
(for $H_x^r$) or $\cos a_p(x+a)\sin b_q(y+b)$ (for $H_y^r$) and is integrated
over the cross-section of the waveguide aperture. Using (\ref{H_field_pot}) 
as well as (\ref{pot_re}) and (\ref{pot_rh}), one can find an expression for
$H_{x,y}^r$.  This results in 
\begin{eqnarray}
\label{Hrx}
\int_{- a}^{a} \int_{- b}^{b}H_x^{r}(x,y,0) \sin a_p (x+a) \cos b_q (y+b) dy dx \nonumber\\
 	= j\frac{ab}{Z_o k_o^2}(k_o b_q A_{pq}^e-k_{pq} A_{pq}^ha_p(1+\delta_{0q}))
\end{eqnarray}
for $p = 1,2,3,...$ and $q = 0,1,2,...$, and
\begin{eqnarray}
\label{Hry}
\int_{- a}^{a} \int_{- b}^{b} H_y^{r}(x,y,0) \cos a_p (x+a) \sin b_q (y+b) dy dx\nonumber\\
	= -j\frac{ab}{Z_o k_o^2}(k_o a_p A_{pq}^e + k_{pq} A_{pq}^hb_q(1+\delta_{0p}))
\end{eqnarray}
for $p = 0,1,2,...$ and $q = 1,2,3,...$ 
where $\delta$ is a delta function of magnitude 1.  The delta function appears 
due to the mismatch between the summations for $\Pi^{re}$ and $\Pi^{rh}$.  
It may also be noted that $p$ ($q$) may go down to zero, even though 
$A_{pq}^e$ ($A_{pq}^h$) does not exist at $p=0$ ($q=0$), due to the fact $a_m$ ($b_q$)
equals zero, thus nulling the term.
The last remaining piece of the equation, the $H_{x,y}^i$ term, 
can be solved using (\ref{pot_ih}) in conjunction with either (\ref{E_field_pot}) 
or (\ref{H_field_pot}), resulting in
\begin{equation}
\label{Hix}
H_x^{i}(x,y,0) =-\frac{k_{10}}{k_0 Z_0} E_y^{i}(x,y,0) = j\frac{a_1 k_{10}}{Z_o k_o^2}A_i \sin a_1(x+a)
\end{equation}
and $H_y^{i}(x,y,0) = E_x^{i}(x,y,0) = 0$.  Using the same method from above, 
equation (\ref{H_boundary}) is multiplied by $\sin a_p(x+a)\cos b_q(y+b)$,
substituting (\ref{Hrx}), (\ref{Hry}), and (\ref{Hix}) for $H^r$ and $H^i$, and is integrated
over the aperture.  This, when combined with (\ref{Htx}), yields the final set of equations:
\begin{eqnarray}
\label{H_x_final}
\sum_{m,n=1}^{\infty} k_{mn}A_{mn}^e[a_mI_1(x,y,p,q) + b_nI_2(x,y,p,q)]\nonumber\cdots\\
						+ \sum_{\stackrel{m,n=0}{m=n\ne0}}^{\infty} k_oA_{mn}^h[b_nI_1(x,y,p,q) - a_mI_2(x,y,p,q)] \nonumber\\
						 =a_1A^i \left[ k_o I_2(1,0,p,q) - 2k_{10}\frac{Z_1}{Z_0}ab\delta_{1p}\delta_{0q}\right] \nonumber\\
						-ab\frac{Z_1}{Z_0}\left[ k_o b_q A_{pq}^e - k_{pq}A_pq^ha_p(1+\delta_{0q})\right] 
\end{eqnarray}
for $p = 1,2,3,...$ and $q = 0,1,2,...$ and 
\begin{eqnarray}
\label{H_y_final}
\sum_{m,n=1}^{\infty} k_{mn}A_{mn}^e[a_mI_3(x,y,p,q) + b_nI_4(x,y,p,q)]\nonumber\\
						+ \sum_{\stackrel{m,n=0}{m=n\ne0}}^{\infty} k_oA_{mn}^h[b_nI_3(x,y,p,q) - a_mI_4(x,y,p,q)] \nonumber\\
						 =a_1 A^i k_0 I_4(1,0,p,q) 
						- ab\frac{Z_1}{Z_0}\left[ k_0 a_p A_{pq}^e + k_{pq}b_qA_{pq}^h(1+\delta_{0p})\right]
\end{eqnarray}
for $p = 0,1,2,...$ and $q = 1,2,3,...$ with
\begin{equation}
\label{I1}
I_1(m,n,p,q) = \frac{1}{4\pi^2}\int_{- \infty}^{\infty} \int_{- \infty}^{\infty}\frac{\xi\eta}{\zeta k_1} C_m^a(-\xi)S_n^b(-\eta)S_p^a(\xi)C_q^b(\eta)d\xi d\eta
\end{equation}

\begin{equation}
\label{I2}
I_2(m,n,p,q) = \frac{1}{4\pi^2}\int_{- \infty}^{\infty} \int_{- \infty}^{\infty}\frac{k_1^2-\xi^2}{\zeta k_1} S_m^a(-\xi)C_n^b(-\eta)S_p^a(\xi)C_q^b(\eta)d\xi d\eta
\end{equation}

\begin{equation}
\label{I3}
I_3(m,n,p,q) = \frac{1}{4\pi^2}\int_{- \infty}^{\infty} \int_{- \infty}^{\infty}\frac{k_1^2-\eta^2}{\zeta k_1} S_m^a(-\xi)C_n^b(-\eta)C_p^a(\xi)S_q^b(\eta)d\xi d\eta
\end{equation}

\begin{equation}
\label{I4}
I_4(m,n,p,q) = \frac{1}{4\pi^2}\int_{- \infty}^{\infty} \int_{- \infty}^{\infty}\frac{\xi\eta}{\zeta k_1} S_m^a(-\xi)C_n^b(-\eta)C_p^a(\xi)S_q^b(\eta) d\xi d\eta
\end{equation}
and

\begin{equation}
\label{Sma}
S_m^a(\xi) = \int_{- a}^{a} \sin a_m(x+a)e^{-j\xi z}dx
\end{equation}

\begin{equation}
\label{Cma}
C_m^a(\xi) = \int_{- a}^{a} \cos a_m(x+a)e^{-j\xi z}dx
\end{equation}

\begin{equation}
\label{Snb}
S_n^b(\eta) = \int_{-b}^{b} \sin b_n(y+b)e^{-j\eta y}dx
\end{equation}

\begin{equation}
\label{Cnb}
C_n^b(\eta) = \int_{-b}^{b} \cos b_n(y+b)e^{-j\eta y}dx.
\end{equation}

From this point, the coefficients $A_{mn}^e$ and $A_{mn}^h$ may be found 
using numerical techniques.

The results (\ref{H_x_final}) and (\ref{H_y_final}) match those
found in \cite{bois}.  It remains to be demonstrated here that (\ref{I1}), (\ref{I2}),
(\ref{I3}), and (\ref{I4}) may be simplified into a form resembling Green's kernel, something
neglected in prior studies.
Additionally, it should be shown that the solution may be found utilizing an alteration 
to the approach herein, reducing the number of steps required to arrive at a solution.

Further work will consider a variety of situations, each growing in 
complexity.  Investigation will proceed to consider a dielectric slab of finite thickness 
backed by a conductor, then moving on to include
lift-off between the probe and the dielectric.  This shall culminate with 
analysis of a conductor-backed, layered, dielectric structure.

With the increase in complication of the problem, so too does its applicability
to practical issues.  In particular, the final situation mentioned, that of the layered 
dielectric structure, arises in may different NDE applications, such as evaluating
composite materials.  The inclusion of a conductor backing enables the modeling of 
structures with a low dielectric constant, such as a radome, allowing for greater 
understanding of the single-sided microwave NDE technique for this particular application.


\section{References}
\bibliographystyle{unsrt} 
\bibliography{open_ended_wg_solution_article}


\end{document}
